calculusgeometry
find a Cartesian equation for the parabola whose
focus is the origin and whose directrix is the line $2x + y = 10$
My attempt :I know one formula that is If $N$ is a vector normal to $L$ and if $P$ is any point on $L $ the distance $d(X, L)$ from any $X$ to $L$ is given by the formula $d(X, L) =\frac{|(X-P). N}{||N||}\tag 1$
Here $L= \{(x,y) | 2+y =10 \}$
I don't know how to colloborate this formula $(1)$ with the given question
I should probably use the fact that $r/d = e$, where $r$ is the distance from the focus to any point $M(x,y)$ of an ellipse. $d$ the distance from $M(x,y)$ to the directrix, and $e$ is the eccentricity.
If you showed your attempt to begin with, we might be able to be a bit more helpful; alas, since you haven't shown it, you'll have to content yourself with the following sketch of a solution.
You know the distance from an arbitrary point $(x,y)$ to the focus $(3,0)$:
$$f=\sqrt{(x-3)^2+(y-0)^2}$$
and you can use the formula for point-line distance (formula 11 here) to get the distance from $(x,y)$ to the line $x+y-1=0$:
$$d=\frac{x+y-1}{\sqrt{1^2+1^2}}$$
from which you use the definition for eccentricity, $\varepsilon=\dfrac{f}{d}$, where $\varepsilon=\dfrac12$.
At once you should obtain an equation with a square root. You can try squaring both sides of the equation and then rearrange things to obtain a two-variable quadratic as usual, but you'll have to justify why the squaring is legal. You should end up with
$$7x^2-2xy+7y^2-46x+2y+71=0$$
(Sketch.) Let $P$ and $Q$ be two points on a parabola, and let $R$ be the point where the respective tangents to the parabola intersect. Let $X$ the midpoint of $PQ$. Then $RX$ is parallel to the axis of symmetry of the parabola (proved below). Draw lines through $P$ and $Q$ parallel to $RX$, and reflect these lines in the respective tangents; the focus $F$ is the intersection of the reflected lines.
To show that $RX$ is parallel to the axis of symmetry: Drop perpendiculars from $P,Q,R$ to the directrix, meeting it at $P',Q',R'$ respectively. As you alluded to, the tangent at $P$ is the perpendicular bisector of the segment $FP'$, and likewise for $Q$ and $FQ'$. So, in $\triangle FP'Q'$, two of the perpendicular bisectors pass through $R$; therefore the third does as well. Since $RR'$ is a line through $R$ and perpendicular to $P'Q'$, it must be the perpendicular bisector, that is, $R'$ is the midpoint of $P'Q'$. By parallels, (the extension of) $RR'$ bisects $PQ$, that is, $RR'$ passes through $X$. So $RX$ is perpendicular to the directrix, as claimed.
Edit: Just for reference, here's what this looks like analytically: The direction of $RX$ is $(2,1)$; reflecting in $RP$ just means exchanging $x$ and $y$ coordinates, so the direction of $PF$ is $(1,2)$, and the line through $P$ in that direction is $2x-y=1$. Reflecting in $RQ$ means negating the $y$-coordinate, so the direction of $QF$ is $(2,-1)$, and the line through $Q$ in that direction is $x+2y=1$. The intersection of these lines is $(\frac35,\frac15)$.
Best Answer
Assume any point $(x,y)$ on the parabola. Given the focal point $(0,0)$ and the directrix $x+2y=10$, we can establish the equal distance from $(x,y)$ to the focus and the directrix as follows
$$\sqrt{x^2+y^2} = \frac{|x+2y-10|}{\sqrt{1^2+2^2}}$$
Simplify to obtain the equation
$$4x^2-4xy+ y^2 +20x+40y-100=0$$